Answered

Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Explore in-depth answers to your questions from a knowledgeable community of experts across different fields. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

A woman of mass 55 kg stands on the rim of a frictionless merry-go-round of radius 2.0m and rotational inertia 1250 kg-m2 that is not moving. She throws a rock of mass 350g horizontally in a direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is 2.0m/s. Afterward, what are (a) the angular speed of the merry-go-round and (b) the linear speed of the woman

Sagot :

okpalawalter8

Answer:

a)   [tex]\omega=9.10*10^{-4}rad/sec[/tex]

b)   [tex]V=1.8217*10^{-3}[/tex]

Explanation:

From the question we are told that:

Mass of woman [tex]M_w=55kg[/tex]

Radius of merry go round [tex]r=2.0m[/tex]

Rotational inertia [tex]i= 1250 kg-m2[/tex]

Mass of rock [tex]M_r=350g \approx 0.350kg[/tex]

Speed of rock [tex]V_r=2.0m/s[/tex]

Tangent angle to the outer edge [tex]\theta=1[/tex]

a)

Generally the equation for conservation of momentum is mathematically given by

[tex]M_r(ucos\theta)r=(I+M_wr^2)\omega[/tex]

[tex]0.350(2.0cos1)(2.0)=(1250+(55)(2.0)^2)\omega[/tex]

[tex]1.3998=1470\omega\\\omega=\frac{1.339}{1470}[/tex]

[tex]\omega=9.10*10^{-4}rad/sec[/tex]

b)

Generally the equation for linear speed V is mathematically given by

[tex]V=r\omega\\V=2.0*9.10*10^{-4}[/tex]

[tex]V=1.8217*10^{-3}[/tex]

Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.